Asked by Dana

A plane is flying at an altitude of 880 m. Sarah is standing on the ground and she observes the angle of elevation to the plane as 67'40'. Twenty-five seconds later the angle of elevation had changed to 24'30'.
(a) How far has the plane flown in that time?
(b) what is the speed to the nearest Km/h of the plane?

Please HELP
I have tried tan67'40'*880 and then the other angle and i subtract but it didn't work
What do i do
Answered by Steve
review the cotangent function. Draw a diagram and you will see that the distance the plane has flown is

880 cot24°30' - 880 cot67°40'
Answered by Reiny
My diagram has Sarah at A
and the first position of the plane at P and the second position at Q
The point on the ground below P is B , and below Q the point is C
triangles APB AND AQC are both right-angled and PB = QC = 880 m

In triangle APB
tan67°40' = 880/AB
AB = 880/tan67°40' ----> I see you multiplied
AB = 361.51 m

in triangle AQC
tan 24°30' = 880/AC
AC = 1930.98 m

distance covered by plane in that time
= (1930.98 - 361.51) m
= 1569.47 m

speed = distance/time
=1569.47 m/25seconds
= 1.56947 km/(25/3600) hrs
= appr 226 km/h
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